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doi 10.1287/opre.1080.0575ec pp. ec1–ec6

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Electronic Companion—“Reducing Delays for Medical Appointments:

A Queueing Approach” by Linda V. Green and Sergei Savin,

Appendix for “Reducing Delays for Medical Appoint-

ments: a Queueing Approach”

Proof of Proposition 1

Below we extend the proof approach outlined in Garcia et al. (2002) to include the state-dependent no-show process. Consider a probability of a service completion in the interval (t, t + ∆) which leaves k = 1, ..., K − 2 patients in the backlog. Four mutually exclusive scenarios can lead to such an event.

First, it is possible that the appointment system was empty at time t− T and 1) a patient requested service between t− T and t − T + ∆t, 2) the patient either has actually arrived for his/her service or exhibited a no-show, but did not reschedule his/her service for a later time, and 3) during the time interval allocated for this patient’s service there were k new appointment requests. The probability of this scenario is p(0, t− T )λN∆t(1 − rγ (k))α(k).

Second, it is possible that the appointment system was empty at time t− T and 1) a patient requested service some time between t − T and t − T + ∆t, 2) the patient exhibited a no-show and rescheduled his/her service for a later time, and 3) during the time interval allocated for this patient’s service there were k−1 new appointment requests.

The probability of this scenario is p(0, t− T )λN∆trγ (k − 1) α(k − 1).

Third, it is possible that 1) a departure that left i = 1, ..., k+1 patients behind occurred between t− T and t − T + ∆t, 2) the first of the patients left behind either shows up or is a no-show who does not reschedule service, and 3) during the period of time allocated for the service of this patient, there are k + 1− i appointment requests. The probability associated with this scenario is D(i, t− T, t − T + ∆t) (1 − rγ(k)) α(k + 1 − i).

Finally, it is also possible that 1) a departure that left i = 1, ..., k patients behind

occurred between t− T and t − T + ∆t, 2) the first of the patients left behind exhibits a no-show and reschedules service, and 3) during the period of time allocated for the service of this patient, there are k− i appointment requests. The probability associated with this scenario is D(i, t− T, t − T + ∆t)rγ(k − 1)α(k − i).

Combining these scenarios, we get

D(k, t, t + ∆t) = p(0, t− T )λN∆t ((1 − rγ (k))α(k) + rγ (k − 1) α(k − 1)) + (1− rγ(k)) α(0)D(k + 1, t − T, t − T + ∆t)

+

Xk i=1

((1− rγ(k)) α(k + 1 − i) + rγ(k − 1)α(k − i))

×D(i, t − T, t − T + ∆t), (A1)

or, equivalently,

d(k, t) = p(0, t− T )λN ((1 − rγ (k))α(k) + rγ (k − 1) α(k − 1)) + (1− rγ(k)) α(0)d(k + 1, t − T )

+

Xk i=1

((1− rγ(k)) α(k + 1 − i) + rγ(k − 1)α(k − i))

×d(i, t − T ). (A2)

Note that for k = 0 we get, following the same argument,

d(0, t) = p(0, t− T )λN(1 − rγ (0))α(0) + (1 − rγ(0)) α(0)d(1, t − T ). (A3)

On the other hand, for k = K− 1 the term D(k + 1, t − T, t − T + ∆t) in (A1) disappears since there can be no departure which leaves behind K customers. Also, if at t− T the system was empty, and there was an arrival of a patient between t− T and t − T + ∆t, any number of subsequent arrivals during the service duration above K − 2 will result in the same future state trajectory. Similarly, if between t− T and t − T + ∆t there was a departure which left i = 1, ..., K − 1 patients behind, then any number of subsequent arrivals during the service duration above K− i − 1 will result in the same future state

trajectory. Thus,

d(K− 1, t) = p(0, t − T )λN

ÃÃ

1−

K−2X

i=0

α(i)

!

(1− rγ(K − 1)) + rγ (K − 2) α(K − 2)

!

+

K−1X

i=1

d(i, t− T )

×

⎛

⎝

⎛

⎝1−

K−1−iX

j=0

α(j)

⎞

⎠(1− rγ(K − 1)) + rγ (K − 2) α(K − 1 − i)

⎞

⎠. (A4)

Finally, (7)-(9) are identical to the results of Proposition 2 in Garcia et al. (2002).

Proof of Proposition 2

The stationary solution to (7)-(9) satisfy

π(k) = d^∗(k)

λN , k = 0, ..., K− 1. (A5)

where

π(k) = lim

t→∞p(k, t), d^∗(k) = lim

t→∞d(k, t), k = 0, ..., K− 1. (A6) Note that

π (K) = 1−

K−1X

k=0

π (k) . (A7)

From (6) it follows that

d^∗(0) = d^∗(0)(1− rγ (0))α(0) + (1 − rγ(0)) α(0)d^∗(1), (A8) d^∗(k) = d^∗(0) ((1− rγ (k))α(k) + rγ (k − 1) α(k − 1)) + (1 − rγ(k)) α(0)d^∗(k + 1)

+

Xk i=1

((1− rγ(k)) α(k + 1 − i) + rγ(k − 1)α(k − i)) d^∗(i), k = 1, ..., K − 2,

d^∗(K− 1) = d^∗(0)

ÃÃ

1−

K−2X

i=0

α(i)

!

(1− rγ(K − 1)) + rγ(K − 2)α(K − 2)

!

+

K−1X

i=1

⎛

⎝

⎛

⎝1−

K−1−iX

j=0

α(j)

⎞

⎠(1− rγ(K − 1)) + rγ(K − 2)α(K − 1 − i)

⎞

⎠d(A9)^∗(i).

Further, defining

f (k) = d^∗(k)

d^∗(0), k = 0, ..., K − 1, (A10) we get

f (0) = 1, f (1) = e^ρ

1− rγ(0) − 1, (A11)

and, using (A9),

f (k + 1) = e^ρ

(1− rγ(k))(f (k)− (1 − rγ (k))α(k) − rγ (k − 1) α(k − 1))

− e^ρ

(1− rγ(k))

à _k X

i=1

((1− rγ(k)) α(k + 1 − i) + rγ(k − 1)α(k − i)) f(i)

!

,

k = 1, ..., K − 2. (A12)

Note that (A10) is equivalent to

π(k) = f (k)π(0), k = 0, 1, ..., K− 1, (A13)

and that, in steady state, the overall arrival rate of patients to the system must be equal to the overall departure rate of patients from the system:

λN

Ã_K−1 X

k=0

π (k)

!

= 1 T

Ã_K X

k=1

(1− rγ (k)) π (k)

!

. (A14)

Note that (A14) is equivalent to

ρπ(0)

Ã_K−1 X

k=0

f (k)

!

=

à _K X

k=0

(1− rγ (k)) π (k)

!

− (1 − rγ(0)) π(0)

= 1− (1 − rγ(0)) π(0)

−r

Ã

π(0)

K−1X

k=0

γ (k) f (k) + γ (K)

Ã

1− π(0)

Ã_K−1 X

k=0

f (k)

!!!

, (A15)

so that

π(0)

Ã

ρ

Ã_K−1 X

k=0

f (k)

!

+ (1− rγ(0)) + r

Ã_K−1 X

k=0

(γ (k)− γ (K))f(k)

!!

= 1− rγ (K) , (A16)

or

π(0) = 1− rγ (K)

1− rγ (K) + ρ^³PK−1i=0 f (i)^´− r^PK−1i=1 (γ (K)− γ (i))f(i) (A17) Combining (A7) with (A13) and (A17), we get

π (0) = 1− rγ (K)

1− rγ (K) + ρ^³PK−1i=0 f (i)^´− r^PK−1i=1 (γ (K)− γ (i))f(i),

π (k) = (1− rγ (K))f(k)

1− rγ (K) + ρ^³PK−1i=0 f (i)^´− r^PK−1i=1 (γ (K)− γ (i))f(i), k = 1, ..., K− 1, π (K) = 1− (1− rγ (K))^³PK−1i=0 f (i)^´

1− rγ (K) + ρ^³PK−1i=0 f (i)^´− r^PK−1i=1 (γ (K)− γ (i))f(i). (A18) Proof of Proposition 3

Using (17), we get for k = 1:

π(1) = ρ

(1− rγ(0))π(0). (A19)

Further, for k = 2 we obtain

³λN + T⁻¹(1− rγ(0))^´π(1) = π (0) λN + π(2)T⁻¹(1− rγ(1)) , (A20)

so that

π(2) =

Ãρ + (1− rγ(0)) (1− rγ(1))

!

π(1)− ρ

(1− rγ(1))π (0)

= ρ

(1− rγ(1))π(1) =

à λN T (1− rγ(1))

!2

π(0). (A21)

Assuming that for all k = 1, ..., M < K− 1

π(k) =

à ρ 1− rγ(k)

!k

π(0), (A22)

we obtain for k = M + 1

(ρ + (1− rγ(M − 1))) π(M) = π (M − 1) ρ + π(M + 1) (1 − rγ(M)) , (A23)

so that

π(M + 1) = 1

(1− rγ(M))((ρ + (1− rγ(M − 1))) π(M) − π (M − 1) ρ)

= ρπ(M )

(1− rγ(M)) =

à ρ

1− rγ(M + 1)

!M+1

π(0). (A24)

Further, for k = K we get

π(K) = ρ

(1− rγ(K − 1))π (K − 1) =

à ρ 1− rγ(K)

!K

π(0), (A25)

Finally, the value of π(0) is obtained using the normalization condition (18):

π(0)

⎛

⎝1 + ρ

(1− γ(1)) + ... +

à ρ 1− γ(K)

!K⎞

⎠= 1 ⇒ π(0) =

⎛

⎝1 +

XK k=1

à ρ 1− γ(k)

!k⎞

⎠

−1

. (A26)